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Posted by on Wednesday, May 13, 2009 at 11:31pm.

the Ksp of CaF2 = 1.46*10^-10 and Ka of HF = 3.5*10^-4 Calculate the pH of a solution in which the solubility of CaF2 = .0100 moles/liter.

so Ksp =[Ca2+][F-]^2 and
Ka =[H3O+][F-]/[HF]

I'm not sure how to continue...

  • chem/solubility - , Thursday, May 14, 2009 at 1:07am

    You need another equation.
    2*solubility = (F^-) + (HF)
    That + Ksp + Ka
    Solve for (H^+).
    I get 0.0579 M = (H^+) or pH of 1.237 which rounds to pH = 1.24.
    Using 1.24 and going through it from the front end give S = 0.0100. I hope this helps.

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